2^x = x^2

Date: 02/13/2002 at 23:26:33
From: Jeremy Kelly
Subject: Exponents
Recently I was trying to do some random math problems to get a little
smarter when I came across something that I couldn't do.
2^x = x^2
I noticed first that there were 2 roots (2 and 4) but there was one
that I didn't know of that was negative. Thinking that this was too
obvious, I switched the problem to 5^x = x^2 and tried to solve. I
took the ln of both sides and tried from there, but I didn't really
know where to go. Using some big 'o' and such, I deduced that x had to
be negative and there was only one Real value for x that worked, (and
graphing proved that), but I didn't know how to find it without
graphing it. If you could help that would be great.
Jeremy Kelly

Date: 02/14/2002 at 09:24:29
From: Doctor Rick
Subject: Re: Exponents
Hi, Jeremy.
I don't know of a way to solve your problem analytically, but I know a
method by which we can find successively better approximations to the
solution. It's called the Newton-Raphson method, and it can be
understood from graphical considerations. Here is an explanation:
Cow Grazing Half the Circle: Newton-Raphson Method
http://mathforum.org/dr.math/problems/julus1.18.98.html
Let's change your problem
2^x = x^2 for what x?
into a function:
f(x) = 2^x - x^2
Now the question is, for what x is f(x) = 0; or, what are the roots of
f(x)? The Newton-Raphson method starts with some first guess, x[0],
and finds the next guess, x[1], by a formula. Then, using this guess,
we apply the same formula to find a new guess, x[2]. We continue until
we're as close as we wish. The formula is
x[i+1] = x[i] - f(x[i])/f'(x[i])
We need f'(x), the derivative of f(x). It is
f'(x) = 2^x * ln(2) - 2x
Thus the formula for our problem is
x[i+1] = x[i] - (2^x[i]-x^2)/(2^x[i]*ln(2)-2x)
You can set this up in a spreadsheet. Then try different first guesses
x[0]. You'll find that the algorithm zeroes in on one of the three
roots, depending on the starting value. If I start with x[0] = 0, I
get the root:
x = -0.766664696
after 5 iterations. You can verify:
2^-0.766664696 = 0.587774756
(-0.766664696)^2 = 0.587774756
If I start with x[0] = 1, I get the root x=2. If I start with
x[0] = 3, I get the root x = 4. You have observed that there are three
roots.
I'll leave the variant with 5^x, and others, up to you.
- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/